The Central Idea
Suppose two points are
\[ P=(x_1,y_1) \]and
\[ Q=(x_2,y_2). \]Coordinate geometry lets us answer two fundamental questions:
- How far apart are \(P\) and \(Q\)?
- Which point lies exactly halfway between them?
The first question leads to the distance formula. The second leads to the midpoint formula.
Distance on a Number Line
Before measuring distance in the plane, recall distance in one dimension.
If two points on a number line have coordinates \(a\) and \(b\), their distance is
\[ \boxed{|b-a|}. \]Absolute value is necessary because distance is nonnegative.
For example, the distance between \(-3\) and \(5\) is
\[ |5-(-3)|=|8|=8. \]Horizontal Distance in the Plane
If two points have the same \(y\)-coordinate, the segment joining them is horizontal.
For
\[ P=(x_1,y) \]and
\[ Q=(x_2,y), \]the distance is simply the horizontal change:
\[ \boxed{|x_2-x_1|}. \]This is the familiar number-line distance applied to the \(x\)-axis direction.
Vertical Distance in the Plane
If two points have the same \(x\)-coordinate, the segment joining them is vertical.
For
\[ P=(x,y_1) \]and
\[ Q=(x,y_2), \]the distance is
\[ \boxed{|y_2-y_1|}. \]Horizontal and Vertical Change
For general points
\[ P=(x_1,y_1),\qquad Q=(x_2,y_2), \]define the coordinate changes
\[ \Delta x=x_2-x_1 \]and
\[ \Delta y=y_2-y_1. \]The symbol \(\Delta\), read “delta,” means change in a quantity.
These changes may be positive, negative, or zero because they retain direction.
Distance Is Different from Coordinate Change
A coordinate change is signed. Distance is not.
If
\[ \Delta x=-6, \]then the horizontal movement is six units to the left. Its horizontal length is
\[ |\Delta x|=6. \]The distinction between signed change and nonnegative length will become especially important when we study slope and direction.
The Right Triangle Hidden Between Two Points
Suppose
\[ P=(x_1,y_1) \]and
\[ Q=(x_2,y_2) \]are not joined by a purely horizontal or vertical segment.
Their horizontal and vertical coordinate changes can be used as the legs of a right triangle:
\[ \text{horizontal leg}=|x_2-x_1|, \] \[ \text{vertical leg}=|y_2-y_1|. \]The segment \(PQ\) is the hypotenuse.
Deriving the Distance Formula
By the Pythagorean relationship,
\[ d^2=(x_2-x_1)^2+(y_2-y_1)^2. \]Distance is nonnegative, so take the nonnegative square root:
\[ \boxed{ d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2} }. \]This is the distance formula in the Cartesian plane.
Why Absolute Values Disappear in the Formula
The horizontal leg could be written
\[ |x_2-x_1|, \]but the Pythagorean calculation squares that length:
\[ |x_2-x_1|^2=(x_2-x_1)^2. \]The same is true vertically.
Squaring removes the sign before the horizontal and vertical contributions are added.
Distance Does Not Depend on Which Point Comes First
Interchanging the points changes the signs of both coordinate differences:
\[ x_1-x_2=-(x_2-x_1) \]and
\[ y_1-y_2=-(y_2-y_1). \]But squaring removes these sign changes:
\[ (x_1-x_2)^2=(x_2-x_1)^2. \]Therefore,
\[ \boxed{d(P,Q)=d(Q,P)}. \]Example: A \(3\)-\(4\)-\(5\) Distance
Find the distance between
\[ P=(1,2) \]and
\[ Q=(4,6). \]Compute the changes:
\[ \Delta x=4-1=3, \] \[ \Delta y=6-2=4. \]Therefore,
\[ d=\sqrt{3^2+4^2} =\sqrt{9+16} =\sqrt{25} =\boxed{5}. \]Example: Negative Coordinates
Find the distance between
\[ A=(-2,3) \]and
\[ B=(4,-5). \]The coordinate differences are
\[ 4-(-2)=6 \]and
\[ -5-3=-8. \]Thus,
\[ d=\sqrt{6^2+(-8)^2} =\sqrt{36+64} =\sqrt{100} =\boxed{10}. \]Example: An Irrational Distance
Find the distance between
\[ (0,0) \]and
\[ (2,3). \]Then
\[ d=\sqrt{2^2+3^2} =\sqrt{13}. \]The exact distance is
\[ \boxed{\sqrt{13}}. \]A decimal approximation is
\[ \sqrt{13}\approx3.606. \]Exact Form and Approximate Form
When a distance contains a square root that does not simplify to a rational number, the radical form is exact.
For example,
\[ \sqrt{13} \]is exact, while
\[ 3.606 \]is an approximation.
Unless a decimal is specifically required, keeping the exact radical form preserves more information.
Simplifying Radical Distances
Suppose a calculation gives
\[ d=\sqrt{72}. \]Factor out the largest perfect-square factor:
\[ \sqrt{72} =\sqrt{36\cdot2} =6\sqrt{2}. \]Therefore,
\[ \boxed{d=6\sqrt{2}}. \]Distance from the Origin
For a point
\[ P=(x,y), \]the origin is
\[ O=(0,0). \]Substituting into the distance formula gives
\[ OP=\sqrt{(x-0)^2+(y-0)^2}. \]Therefore,
\[ \boxed{OP=\sqrt{x^2+y^2}}. \]Distance as a Geometric Constraint
Distance formulas do more than measure completed diagrams. They can also describe sets of possible points.
If a point \((x,y)\) must remain exactly \(r\) units from the origin, then
\[ \sqrt{x^2+y^2}=r. \]Squaring gives
\[ x^2+y^2=r^2. \]This is the standard equation of a circle centered at the origin.
Thus, geometric loci can emerge directly from distance conditions.
What Is a Midpoint?
The midpoint of a segment is the point on the segment that divides it into two congruent segments.
If \(M\) is the midpoint of \(\overline{PQ}\), then
\[ PM=MQ. \]In coordinates, the midpoint is found by averaging the corresponding coordinates of the endpoints.
The Midpoint Formula
If
\[ P=(x_1,y_1) \]and
\[ Q=(x_2,y_2), \]then their midpoint is
\[ \boxed{ M= \left( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} \right) }. \]The \(x\)-coordinates are averaged with each other, and the \(y\)-coordinates are averaged with each other.
Why the Midpoint Formula Uses Averages
On a number line, the number halfway between \(a\) and \(b\) is their arithmetic mean:
\[ \frac{a+b}{2}. \]A Cartesian midpoint applies this one-dimensional idea independently in two perpendicular directions.
Horizontally, the midpoint lies halfway between \(x_1\) and \(x_2\). Vertically, it lies halfway between \(y_1\) and \(y_2\).
Example: Finding a Midpoint
Find the midpoint of
\[ A=(2,4) \]and
\[ B=(8,10). \]Average the \(x\)-coordinates:
\[ \frac{2+8}{2}=5. \]Average the \(y\)-coordinates:
\[ \frac{4+10}{2}=7. \]Therefore,
\[ \boxed{M=(5,7)}. \]Example: Midpoint with Negative Coordinates
Find the midpoint of
\[ P=(-5,3) \]and
\[ Q=(1,-7). \]Then
\[ M= \left( \frac{-5+1}{2}, \frac{3+(-7)}{2} \right). \]Simplify:
\[ M= \left( \frac{-4}{2}, \frac{-4}{2} \right). \]Thus,
\[ \boxed{M=(-2,-2)}. \]Midpoints Can Have Fractional Coordinates
Consider the endpoints
\[ (1,2) \]and
\[ (4,8). \]Their midpoint is
\[ \left( \frac{1+4}{2}, \frac{2+8}{2} \right) = \left( \frac{5}{2}, 5 \right). \]There is no requirement that midpoint coordinates be integers.
The Midpoint Is Symmetric in the Endpoints
Switching the endpoint order does not change the midpoint:
\[ \frac{x_1+x_2}{2} = \frac{x_2+x_1}{2}. \]Likewise,
\[ \frac{y_1+y_2}{2} = \frac{y_2+y_1}{2}. \]Therefore the midpoint of \(P\) and \(Q\) is the same as the midpoint of \(Q\) and \(P\).
Checking a Midpoint with Distance
If a point is truly the midpoint, it should be equally distant from both endpoints.
For
\[ P=(0,0),\qquad Q=(6,8), \]the midpoint is
\[ M=(3,4). \]Then
\[ PM=\sqrt{3^2+4^2}=5 \]and
\[ MQ=\sqrt{3^2+4^2}=5. \]The equal distances confirm the midpoint calculation.
A Midpoint Is More Than an Equidistant Point
Being equally distant from two endpoints is necessary for a midpoint, but it is not sufficient by itself.
Many points can be equally distant from the endpoints of a segment. In fact, every point on the segment's perpendicular bisector has that property.
A midpoint must satisfy two conditions:
- it lies on the segment joining the endpoints;
- it divides that segment into two equal parts.
Recovering an Unknown Endpoint
The midpoint formula can be used backward.
Suppose one endpoint is
\[ A=(2,5), \]the midpoint is
\[ M=(6,1), \]and the other endpoint is
\[ B=(x,y). \]Then
\[ \frac{2+x}{2}=6, \qquad \frac{5+y}{2}=1. \]Solving gives
\[ x=10,\qquad y=-3. \]Therefore,
\[ \boxed{B=(10,-3)}. \]A Shortcut for Recovering the Other Endpoint
From
\[ M_x=\frac{x_1+x_2}{2}, \]solve for \(x_2\):
\[ x_2=2M_x-x_1. \]Similarly,
\[ y_2=2M_y-y_1. \]Therefore, if one endpoint and the midpoint are known, the opposite endpoint can be found by
\[ \boxed{ (x_2,y_2)= (2M_x-x_1,\ 2M_y-y_1) }. \]Midpoint as Half of a Displacement
Another way to reach the midpoint from
\[ P=(x_1,y_1) \]is to move half of the displacement from \(P\) to \(Q\).
The full displacement is
\[ (x_2-x_1,\ y_2-y_1). \]Half of it is
\[ \left( \frac{x_2-x_1}{2}, \frac{y_2-y_1}{2} \right). \]Adding that half-displacement to \(P\) produces the same midpoint formula.
Distance and Midpoint Work Together
Distance measures the length of a segment. Midpoint divides that same segment into two equal pieces.
If
\[ PQ=d, \]and \(M\) is the midpoint of \(\overline{PQ}\), then
\[ \boxed{PM=MQ=\frac{d}{2}}. \]This relationship provides a useful geometric check.
Using Distance to Test Congruent Segments
Two segments are congruent when they have the same length.
Suppose
\[ A=(0,0),\quad B=(3,4),\quad C=(2,1),\quad D=(5,5). \]Then
\[ AB=\sqrt{3^2+4^2}=5 \]and
\[ CD=\sqrt{(5-2)^2+(5-1)^2} =\sqrt{3^2+4^2} =5. \]Therefore,
\[ \boxed{\overline{AB}\cong\overline{CD}}. \]Using Distance to Classify a Triangle
Coordinates can verify whether a triangle has equal sides.
Consider
\[ A=(0,0),\qquad B=(4,0),\qquad C=(2,3). \]We have
\[ AC=\sqrt{2^2+3^2}=\sqrt{13} \]and
\[ BC=\sqrt{(-2)^2+3^2}=\sqrt{13}. \]Since \(AC=BC\), the triangle is isosceles.
Coordinate calculation has verified a geometric classification.
Using Midpoints to Test Diagonal Bisection
Many geometric claims can be translated into midpoint statements.
If two diagonals have the same midpoint, then they bisect each other.
For example, suppose diagonal endpoints are
\[ A=(0,0),\quad C=(6,4) \]and
\[ B=(2,5),\quad D=(4,-1). \]The midpoint of \(\overline{AC}\) is
\[ (3,2), \]and the midpoint of \(\overline{BD}\) is also
\[ (3,2). \]Therefore the diagonals bisect each other.
Squared Distance as a Useful Shortcut
Sometimes we only need to compare distances rather than calculate their square roots.
Define the squared distance
\[ d^2=(x_2-x_1)^2+(y_2-y_1)^2. \]Because the square-root function is increasing for nonnegative inputs, comparing \(d^2\) values gives the same ordering as comparing \(d\) values.
This can simplify calculations in coordinate proofs and algorithms.
Example: Compare Distances Without Radicals
From the origin, compare the distances to
\[ A=(3,7) \]and
\[ B=(5,5). \]Their squared distances are
\[ OA^2=3^2+7^2=58 \]and
\[ OB^2=5^2+5^2=50. \]Since
\[ 58>50, \]we know
\[ \boxed{OA>OB} \]without evaluating either square root.
Three-Dimensional Preview
The same ideas extend naturally to three dimensions.
For
\[ P=(x_1,y_1,z_1) \]and
\[ Q=(x_2,y_2,z_2), \]the distance is
\[ \sqrt{ (x_2-x_1)^2+ (y_2-y_1)^2+ (z_2-z_1)^2 }. \]The midpoint is
\[ \left( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}, \frac{z_1+z_2}{2} \right). \]A later lesson will develop three-dimensional coordinates in full.
Common Error: Forgetting to Subtract Coordinates
The distance formula uses coordinate differences, not the coordinates themselves.
Between
\[ (2,3) \]and
\[ (7,11), \]the relevant changes are
\[ 7-2=5 \]and
\[ 11-3=8. \]The formula does not use \(2^2+3^2+7^2+11^2\).
Common Error: Mishandling a Negative Coordinate
If
\[ x_1=-4,\qquad x_2=3, \]then
\[ x_2-x_1=3-(-4)=7. \]Subtracting a negative becomes addition.
Writing \(3-4=-1\) would lose the sign of the original coordinate.
Common Error: Adding Coordinate Differences Before Squaring
The distance formula is
\[ \sqrt{(\Delta x)^2+(\Delta y)^2}, \]not
\[ \sqrt{(\Delta x+\Delta y)^2}. \]Horizontal and vertical changes are perpendicular components. Their squares are added according to the Pythagorean relationship.
Common Error: Forgetting the Square Root
If
\[ (x_2-x_1)^2+(y_2-y_1)^2=25, \]then \(25\) is the squared distance.
The actual distance is
\[ \boxed{\sqrt{25}=5}. \]Common Error: Averaging All Four Coordinates Together
The midpoint formula does not combine every coordinate into one average.
For
\[ (2,6) \]and
\[ (8,10), \]average the \(x\)-coordinates together and the \(y\)-coordinates together:
\[ \left( \frac{2+8}{2}, \frac{6+10}{2} \right) = \boxed{(5,8)}. \]Common Error: Dividing Only One Endpoint by Two
The midpoint coordinate
\[ \frac{x_1+x_2}{2} \]means the sum of the two endpoint coordinates is divided by \(2\).
Parentheses can make the structure explicit:
\[ \frac{(x_1+x_2)}{2}. \]Do not interpret it as
\[ x_1+\frac{x_2}{2}. \]A Reliable Procedure for Distance
- Label the endpoints \((x_1,y_1)\) and \((x_2,y_2)\).
- Compute the horizontal difference \(x_2-x_1\).
- Compute the vertical difference \(y_2-y_1\).
- Square both differences.
- Add the squares.
- Take the nonnegative square root.
- Simplify the radical if possible.
- Approximate only when a decimal value is needed.
A Reliable Procedure for Midpoint
- Identify the two \(x\)-coordinates.
- Add them and divide by \(2\).
- Identify the two \(y\)-coordinates.
- Add them and divide by \(2\).
- Write the two averages as an ordered pair.
- Check that the point visually and numerically lies halfway between the endpoints.
Practice 1: Horizontal Distance
Find the distance between
\[ (-4,3) \]and
\[ (7,3). \]Practice 2: Vertical Distance
Find the distance between
\[ (5,-6) \]and
\[ (5,2). \]Practice 3: Distance Formula
Find the distance between
\[ (1,1) \]and
\[ (6,13). \]Practice 4: Radical Distance
Find the exact distance between
\[ (-1,2) \]and
\[ (3,5). \]Practice 5: Distance from the Origin
Find the distance from
\[ (-8,15) \]to the origin.
Practice 6: Midpoint
Find the midpoint of
\[ (2,-3) \]and
\[ (10,5). \]Practice 7: Fractional Midpoint
Find the midpoint of
\[ (-4,1) \]and
\[ (3,6). \]Practice 8: Missing Endpoint
One endpoint is
\[ A=(3,-2), \]and the midpoint is
\[ M=(7,4). \]Find the other endpoint \(B\).
Practice 9: Compare Distances
Which point is closer to the origin:
\[ A=(5,6) \]or
\[ B=(2,8)? \]Practice 10: Identify the Error
A student finds the midpoint of \((2,8)\) and \((10,4)\) by writing
\[ \left( \frac{2+8}{2}, \frac{10+4}{2} \right) =(5,7). \]Explain the error and give the correct midpoint.
Practice Answers
Practice 1
The points have the same \(y\)-coordinate, so use horizontal distance:
\[ |7-(-4)|=\boxed{11}. \]Practice 2
The points have the same \(x\)-coordinate:
\[ |2-(-6)|=\boxed{8}. \]Practice 3
\[ d=\sqrt{(6-1)^2+(13-1)^2} =\sqrt{5^2+12^2} =\sqrt{169} =\boxed{13}. \]Practice 4
\[ d=\sqrt{(3-(-1))^2+(5-2)^2} =\sqrt{4^2+3^2} =\boxed{5}. \]Practice 5
\[ d=\sqrt{(-8)^2+15^2} =\sqrt{64+225} =\sqrt{289} =\boxed{17}. \]Practice 6
\[ M= \left( \frac{2+10}{2}, \frac{-3+5}{2} \right) = \boxed{(6,1)}. \]Practice 7
\[ M= \left( \frac{-4+3}{2}, \frac{1+6}{2} \right) = \boxed{ \left( -\frac12, \frac72 \right) }. \]Practice 8
Use the midpoint equations:
\[ \frac{3+x}{2}=7, \qquad \frac{-2+y}{2}=4. \]Therefore,
\[ x=11,\qquad y=10, \]so
\[ \boxed{B=(11,10)}. \]Practice 9
Compare squared distances:
\[ OA^2=5^2+6^2=61, \] \[ OB^2=2^2+8^2=68. \]Since \(61<68\),
\[ \boxed{A\text{ is closer to the origin}}. \]Practice 10
The student paired an \(x\)-coordinate with a \(y\)-coordinate. Corresponding coordinate types must be averaged together:
\[ M= \left( \frac{2+10}{2}, \frac{8+4}{2} \right) = \boxed{(6,6)}. \]What to Remember
For points
\[ P=(x_1,y_1),\qquad Q=(x_2,y_2), \]the distance formula is
\[ \boxed{ d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2} }. \]The midpoint formula is
\[ \boxed{ M= \left( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} \right) }. \]- distance comes from the Pythagorean relationship;
- distance is always nonnegative;
- coordinate differences measure horizontal and vertical change;
- midpoint coordinates are averages of corresponding endpoint coordinates;
- exact radicals should generally be preserved unless approximation is required;
- distance and midpoint formulas can verify geometric claims algebraically.
Where We Go Next
Distance measures how much separation exists between two coordinate points. The next question is how a segment is oriented and how rapidly its vertical position changes relative to its horizontal position.
The next lesson, Slope, Direction, and Inclination, develops signed rates of change, line direction, and the geometric meaning of slope.